But the author's analysis is faulty - he assumes an impossible condition in the final state for his illustrative process. That impossible condition then gives rise to the "startling" behavior of the entropy.

With respect to the details of Belandria's process: the surroundings deliver 14.055 kJ of work to subsystem A; the energy equivalent of that work flows into subsystem B and increases the temperature of B from 373 K to 1500 K.

The point the author failed to appreciate is that the path specified for the work interaction of 14.055 kJ uniquely determines the final pressure, P_A(final), of the gas in subsystem A.

Examples

Example 1

Let the wall of subsystem A be pierced by a thermally nonconducting rod that presses against a friction pad in subsystem A. Rotate the rod against the friction pad, doing 14.055 kJ of work in the process. The energy equivalent of the dissipated work finds its way into subsystem B and raises the temperature of B to 1500 K. The pressure of the gas in subsystem A remains at 1 atm.

Example 2

Compress gas A by using a constant external pressure of 5 atm, and solve for Delta V_A from the relationship -P(external)Delta V_A = 14.055 kJ. The final pressure, P_A(final), turns out to be 1.29 atm.

Example 3

Let a series of constant external pressuressay, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0 atmeach bring about the same change in volume Delta V_A'. Solve for Delta V_A' from -(1.5 + 2.0 + 2.5 + 3.0 + 3.5 + 4.0) Delta V_A' = 14.055 kJ. Then Delta V_A = 6 Delta V_A' and P_A(final) = 1.69 atm.

Example 4

Compress gas A reversibly and isothermally: 14.055 kJ = R ¥ 1500 ln (P_A(final) / 1 atm) and P_A(final) = 3.086 atm.

This is the end of the line. For the given conditions, 3.086 atm is the maximum pressure that can be attained in subsystem A. The range of allowed values for P_A(final) is thus 1 £ P_A(final) £ 3.086 atm.

By assuming a pressure of 4 atm for P_A(final), the author has imposed an impossible condition on the processthereby causing the entropy to behave in the "weird" way he describes.